Here's the proof for the partial sum:
S=:\sum_{k=1}^{n}\frac{1}{k(k+1)(k+2)}=\frac{1}{4}-\frac{1}{2(n+1)(n+2)}(1)
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Use partial fractions to rewrite the initial partial sum as:
S=\sum_{k=1}^{n}\frac{1}{k(k+1)(k+2)}=\sum_{k=1}^{n}\frac{1}{2k}-\sum_{k=1}^{n}\frac{1}{k+1}+\sum_{k=1}^{n}\frac{1}{2(k+2)} (2)
According to Maple each of the three sums is:
\sum_{k=1}^{n}\frac{1}{2k}=\frac{1}{2}\psi(n+1)+\frac{1}{2}\gamma (3)
\sum_{k=1}^{n} \frac{-1}{k+1}=-\psi(n+2)+1-\gamma (4)
\sum_{k=2}^{n} \frac{1}{2(k+2)}=\frac{1}{2}\psi(n+3)-\frac{3}{4}+\frac{1}{2}\gamma(5)
Add (3)->(5) and equate with (2):
S=\frac {1}{4}+\frac{1}{2}\psi(n+1)-\psi(n+2)+\frac{1}{2}\psi(n+3)(6)
Now,use the property of "psi":
\psi(z+1)=\psi(z)+\frac{1}{z} (7) for z\neq 0
to write:
\psi(n+2)=\psi(n+1)+\frac{1}{n+1} (8)
\psi(n+3)=\psi(n+2)+\frac{1}{n+2}=\psi(n+1)+\frac{1}{n+1}+\frac{1}{n+2} (9)
to rewrite the "3-psi" term from (6) simply as:
-\frac{1}{2(n+1)(n+2)} (10)
Then add (10) to 1/4 from (6) to get the equality you were supposed to prove...
In the initial equality set n\rightarrow +\infty to get the answer \frac{1}{4}
Daniel.
